Verify the Rodrigues formula of the Legendre polynomials

Using product rule in reverse on the first two terms,
$$(1-w^2)\frac{d^{l+1}}{dw^{l+1}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$
$$(1-w^2)\frac{d^{l+1}}{dw^{l+1}}(w^2-1)^l-2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$

Using product rule in reverse on the first two terms,
$$(1-w^2)\frac{d^{l}}{dw^{l}}(w^2-1)^l+2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$
$$\big[(1+2w-w^2)+l(l+1)\big]\frac{d^l}{dw^l}(w^2-1)^l$$

I think it can be easily solved with the help of Leibniz rule for the differentiation of a product of two functions
$$
\frac{d^n}{dx^n}(f(x)g(x)) = \sum_{k=0}^n C(n,k) \frac{d^{n-k}f}{dx^{n-k}} \frac{d^k g}{dx^n}
$$
In the terms which contain differentiation ##l+1## or ##l+2## times, differentiate the function 1 or 2 times first.

I think it can be easily solved with the help of Leibniz rule for the differentiation of a product of two functions
$$
\frac{d^n}{dx^n}(f(x)g(x)) = \sum_{k=0}^n C(n,k) \frac{d^{n-k}f}{dx^{n-k}} \frac{d^k g}{dx^n}
$$
In the terms which contain differentiation ##l+1## or ##l+2## times, differentiate the function 1 or 2 times first.